Binary Tree Paths

// Accepted solution for LeetCode #257: Binary Tree Paths
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private List<String> ans = new ArrayList<>();
    private List<String> t = new ArrayList<>();

    public List<String> binaryTreePaths(TreeNode root) {
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        t.add(root.val + "");
        if (root.left == null && root.right == null) {
            ans.add(String.join("->", t));
        } else {
            dfs(root.left);
            dfs(root.right);
        }
        t.remove(t.size() - 1);
    }
}

// Accepted solution for LeetCode #257: Binary Tree Paths
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func binaryTreePaths(root *TreeNode) (ans []string) {
	t := []string{}
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		t = append(t, strconv.Itoa(root.Val))
		if root.Left == nil && root.Right == nil {
			ans = append(ans, strings.Join(t, "->"))
		} else {
			dfs(root.Left)
			dfs(root.Right)
		}
		t = t[:len(t)-1]
	}
	dfs(root)
	return
}

# Accepted solution for LeetCode #257: Binary Tree Paths
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            t.append(str(root.val))
            if root.left is None and root.right is None:
                ans.append("->".join(t))
            else:
                dfs(root.left)
                dfs(root.right)
            t.pop()

        ans = []
        t = []
        dfs(root)
        return ans

// Accepted solution for LeetCode #257: Binary Tree Paths
struct Solution;
use rustgym_util::*;

struct Path {
    stack: Vec<i32>,
}

impl ToString for Path {
    fn to_string(&self) -> String {
        let s: Vec<String> = self.stack.iter().map(|x| x.to_string()).collect();
        s.join("->")
    }
}

trait Preorder {
    fn preorder(&self, path: &mut Path, v: &mut Vec<String>);
}

impl Preorder for TreeLink {
    fn preorder(&self, path: &mut Path, v: &mut Vec<String>) {
        if let Some(node) = self {
            let node = node.borrow();
            path.stack.push(node.val);
            if node.left.is_none() && node.right.is_none() {
                v.push(path.to_string());
            }
            if node.left.is_some() {
                node.left.preorder(path, v);
            }
            if node.right.is_some() {
                node.right.preorder(path, v);
            }
            path.stack.pop();
        }
    }
}

impl Solution {
    fn binary_tree_paths(root: TreeLink) -> Vec<String> {
        let mut path = Path { stack: vec![] };
        let mut res = vec![];
        root.preorder(&mut path, &mut res);
        res
    }
}

#[test]
fn test() {
    let root = tree!(1, tree!(2, None, tree!(5)), tree!(3));
    let paths: Vec<String> = vec_string!["1->2->5", "1->3"];
    assert_eq!(Solution::binary_tree_paths(root), paths);
}

// Accepted solution for LeetCode #257: Binary Tree Paths
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function binaryTreePaths(root: TreeNode | null): string[] {
    const ans: string[] = [];
    const t: number[] = [];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        t.push(root.val);
        if (!root.left && !root.right) {
            ans.push(t.join('->'));
        } else {
            dfs(root.left);
            dfs(root.right);
        }
        t.pop();
    };
    dfs(root);
    return ans;
}